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A baby’s mouth is 30 cm from her father’s ear and 1.50 m from her mother’s ear. What is the difference between the sound intensity levels heard by the father and by the mother?

Short Answer

Expert verified

The difference in the sound intensity heard by the father and by the mother is 14 dB

Step by step solution

01

Concept of the difference in the sound intensity level

The sound intensity level is given by β2-β1=10dBlogl2l1Where, β2-β1is the difference in the sound intensity level, l2is the intensity of the sound heard by the father, l1is the intensity of the sound heard by the mother.

02

Calculate the ratio of the sound intensity heard by the father to the sound heard by the mother with respect distance

The ratio of the sound intensity heard by the father to the sound heard by the mother

with respect distance is, l2l1=r12r22Where, r1is the distance of the mother from the baby,r2 is the distance of the father from the baby.

Substitute 30 cm for r2and 1.5 m for r1in the above equation to findl2l1

l2l1=1.5m100cm21m30cm2=25

Thus, ratio of the sound intensity heard by the father to the sound heard by the mother with respect distance is 25

03

To find the difference in the sound intensity level

Substitute 25 forl2l1 to findβ2-β1

β2-β1=10dBlog25=10dB1.4=14dB

Therefore, the difference in the sound intensity heard by the father and by the mother is 14dB

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